Applications of Derivatives 2
", "licence": "None specified"}, "duration": 0, "percentPass": 0, "showQuestionGroupNames": false, "shuffleQuestionGroups": false, "showstudentname": true, "question_groups": [{"name": "Group", "pickingStrategy": "all-ordered", "pickQuestions": 1, "questionNames": ["", "", "", "", "", ""], "variable_overrides": [[], [], [], [], [], []], "questions": [{"name": "Maria's copy of Maclaurin series (three terms)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "Find the first 3 terms in the MacLaurin series for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "You are asked to find the first 3 terms in the MacLaurin series for $f(x)=(\\simplify[all]{{a}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$ and $x$ is near $0$.
", "advice": "The first three terms in the MacLaurin series are given by $a+bx+cx^2$ where $\\displaystyle a=f(0),\\;\\;b=f'(0),\\;\\;c=\\frac{f''(0)}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(0)=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(0)=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(0)}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the MacLaurin series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*x+{tm2}/{2*a^2*n^2}*x^2} \\]
First 3 terms = ?
\nInput the first three terms in the Taylor series in the form $a+b(x-\\var{c})+c(x-\\var{c})^2$ for suitable coefficients $a,\\;b$ and $c$.
\nInput coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.
", "answer": "{tm0}+{tm1}/{a*n}*x+{tm2}/{2*a^2*n^2}*x^2", "answerSimplification": "all,fractionNumbers,!collectNumbers", "showPreview": true, "checkingType": "absdiff", "checkingAccuracy": 1e-06, "failureRate": 1, "vsetRangePoints": 5, "vsetRange": [0, 1], "checkVariableNames": false, "valuegenerators": [{"name": "x", "value": ""}]}]}, {"name": "Maria's copy of Taylor series (three terms)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Newcastle University Mathematics and Statistics", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/697/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "tags": [], "metadata": {"description": "Find the first 3 terms in the Taylor series at $x=c$ for $f(x)=(a+bx)^{1/n}$ i.e. up to and including terms in $x^2$.
", "licence": "Creative Commons Attribution 4.0 International"}, "statement": "You are asked to find the first 3 terms in the Taylor series at $x=\\var{c}$ for $f(x)=(\\simplify[all]{{a-b*c}+{b}*x})^{1/\\var{n}}$ i.e. up to terms in $x^2$.
", "advice": "The first three terms in the Taylor series are given by $\\simplify[all]{a+b(x-{c})+c(x-{c})^2}$ where $\\displaystyle a=f(\\var{c}),\\;\\;b=f'(\\var{c}),\\;\\;c=\\frac{f''(\\var{c})}{2}$
For this example,
\\[\\begin{eqnarray*} f'(x)&=&\\simplify[all,fractionNumbers]{{b}/{n}*({a-b*c}+{b}x)^(-{n-1}/{n})}\\\\ f''(x)&=&\\simplify[all,fractionNumbers]{-{b^2*(n-1)}/{n^2}*({a-b*c}+{b}x)^(-{2*n-1}/{n})} \\end{eqnarray*} \\]
and so we get:
\\[\\begin{eqnarray*} a&=&f(\\var{c})=\\simplify[all]{{a}^(1/{n})={tm0}}\\\\ b&=&f'(\\var{c})=\\simplify[all,fractionNumbers]{{tm1}/{a*n}}\\\\ c&=&\\frac{f''(\\var{c})}{2}=\\simplify[all,fractionNumbers]{{tm2}/{2*a^2*n^2}} \\end{eqnarray*}\\]
Hence the first three terms of the Taylor series are:
\\[\\simplify[all,fractionNumbers,!collectNumbers]{{tm0}+{tm1}/{a*n}*(x-{c})+{tm2}/{2*a^2*n^2}*(x-{c})^2} \\]
Input the first three terms in the Taylor series in the form $a+b(x-\\var{c})+c(x-\\var{c})^2$ for suitable coefficients $a,\\;b$ and $c$.
\n
Input coefficients as fractions, not as decimals. Also do not use factorials in your answer. For example, input 6 rather than 3!.
Do not input factorials or decimals in the Taylor series.
"}, "valuegenerators": [{"name": "x", "value": ""}]}]}, {"name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form infinity/infinity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "advice": "As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)=\\infty$, $\\lim_{x\\rightarrow a}g(x)=\\infty$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:
\n\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]
\n(If this is still an indeterminate form we can (of course) repeat the procedure)
\n\nSo by applying L'Hospital's rule we now need to determine what {choice2[3]} approaches as {choice2[0]}. But this is again an indeterminate form so we repeatedly apply L'Hospital's rule until it is not an indeterminate form and we arrive at needing to determine what {choice2[4]} approaches as {choice2[0]}. Now that this is no longer an indeterminate form, it should be clear that the limit is $\\var{choice2[-1]}$.
\n", "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "log_power", "power_log", "exp_power", "power_exp", "exp_log", "log_exp", "log_root", "root_log", "choice1", "choice2"], "parts": [{"type": "gapfill", "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"correctAnswerFraction": false, "type": "numberentry", "minValue": "choice2[-1]", "allowFractions": false, "mustBeReducedPC": 0, "scripts": {}, "mustBeReduced": false, "variableReplacementStrategy": "originalfirst", "notationStyles": ["plain", "en", "si-en"], "maxValue": "choice2[-1]", "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "marks": 1, "correctAnswerStyle": "plain"}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "marks": 0, "prompt": "As {choice2[0]}, {choice2[2]} approaches [[0]]
\nNote: infinity is simply entered by typing infinity
\n"}], "variables": {"power_exp": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}e^({pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd}*{pb}x^{pb-1})/({pc}*{pa}*e^({pa}x+{nb}))}\\$','\\$\\\\simplify[!othernumbers]{({pd}*{pb}!)/({pc}*{pa}^{pb}*e^({pa}x+{nb}))}\\$',0], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}e^({pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({-pd}*{pb}x^{pb-1})/({pc}*{pa}*e^({pa}x+{nb}))}\\$','\\$\\\\simplify[!othernumbers]{({-pd}*{pb}!)/({pc}*{pa}^{pb}*e^({pa}x+{nb}))}\\$',0], \n\n\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}e^({-pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({-pd}*{pb}x^{pb-1})/({pc}*{-pa}e^({-pa}x+{nb}))}\\$','\\$\\\\simplify[!othernumbers]{({-pd}*{pb}!)/({pc}*{-pa}^{pb}*e^({-pa}x+{nb}))}\\$',0],\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}e^({-pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd}*{pb}x^{pb-1})/({pc}*{-pa}e^({-pa}x+{nb}))}\\$','\\$\\\\simplify[!othernumbers]{({pd}*{pb}!)/({pc}*{-pa}^{pb}*e^({-pa}x+{nb}))}\\$',0]\n]\n", "name": "power_exp", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
"}, "exp_power": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pc}e^({pa}x+{nb})+{nc})/({pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({pc}*{pa}*e^({pa}x+{nb}))/({pd}*{pb}x^{pb-1})}\\$','\\$\\\\simplify[!othernumbers]{({pc}*{pa}^{pb}*e^({pa}x+{nb}))/({pd}*{pb}!)}\\$',infinity], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pc}e^({pa}x+{nb})+{nc})/({pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({-pc}*{pa}*e^({pa}x+{nb}))/({pd}*{pb}x^{pb-1})}\\$','\\$\\\\simplify[!othernumbers]{({-pc}*{pa}^{pb}*e^({pa}x+{nb}))/({pd}*{pb}!)}\\$',-infinity], \n\n\n\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{-\\\\infty}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({pc}e^({-pa}x+{nb})+{nc})/({-pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({pc}*{-pa}e^({-pa}x+{nb}))/({-pd}*{pb}x^{pb-1})}\\$','\\$\\\\simplify[!othernumbers]{({pc}*{-pa}^{pb}*e^({-pa}x+{nb}))/({-pd}*{pb}!)}\\$',if(mod(pb,2)=0,-infinity,infinity)],\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\frac{\\\\infty}{-\\\\infty}\\$'), '\\$\\\\simplify{({-pc}e^({-pa}x+{nb})+{nc})/({-pd}x^{pb}+{nd})}\\$','\\$\\\\simplify{({pc}*{pa}e^({-pa}x+{nb}))/({-pd}*{pb}x^{pb-1})}\\$','\\$\\\\simplify[!othernumbers]{({pc}*{pa}^{pb}*e^({-pa}x+{nb}))/({-pd}*{pb}!)}\\$',if(mod(pb,2)=0,infinity,-infinity)]\n]\n", "name": "exp_power", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
"}, "log_exp": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}ln({pb}x+{nb}))/({pc}e^({pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd*pb}/({pb}x+{nb}))/({pc}*{pa}*e^({pa}x+{nb}))={pd*pb}/({pc}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))}\\$',0], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}ln({pb}x+{nb}))/({pc}e^({pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({-pd*pb}/({pb}x+{nb}))/({pc}*{pa}*e^({pa}x+{nb}))={-pd*pb}/({pc}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))}\\$',0], \n\n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({pd}ln({-pb}x+{nb}))/({pc}e^({-pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd*-pb}/({-pb}x+{nb}))/({pc}*{-pa}*e^({-pa}x+{nb}))={pd*-pb}/({pc}*{-pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))}\\$',0],\n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({-pd}ln({-pb}x+{nb}))/({pc}e^({-pa}x+{nb})+{nc})}\\$','\\$\\\\simplify{({pd*pb}/({-pb}x+{nb}))/({pc}*{-pa}*e^({-pa}x+{nb}))={pd*pb}/({pc}*{-pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))}\\$',0]\n]", "name": "log_exp", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
"}, "pc": {"templateType": "anything", "definition": "pl[2]", "name": "pc", "group": "Ungrouped variables", "description": ""}, "choice2": {"templateType": "anything", "definition": "switch(choice1=0,random(log_power),choice1=1,random(power_log),choice1=2,random(exp_power),choice1=3,random(power_exp),choice1=4,random(exp_log),choice1=5,random(log_exp),choice1=6,random(log_root),choice1=7,random(root_log),'')", "name": "choice2", "group": "Ungrouped variables", "description": ""}, "nc": {"templateType": "anything", "definition": "nl[2]", "name": "nc", "group": "Ungrouped variables", "description": ""}, "pb": {"templateType": "anything", "definition": "pl[1]", "name": "pb", "group": "Ungrouped variables", "description": ""}, "power_log": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pd*pb}x^{pb-1})/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})*({pd*pb}x^{pb-1}))/({pc*pa})}\\$',infinity], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({-pd*pb}x^{pb-1})/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})*({-pd*pb}x^{pb-1}))/({pc*pa})}\\$',-infinity], \n\n\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({-pd}x^{pb}+{nd})/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pd*pb}x^{pb-1})/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})*({-pd*pb}x^{pb-1}))/({-pc*pa})}\\$',if(mod(pb,2)=0,-infinity,infinity)],\n['\\$x\\\\rightarrow-\\\\infty\\$',if(mod(pb,2)=0,'\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$'), '\\$\\\\simplify{({pd}x^{pb}+{nd})/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({pd*pb}x^{pb-1})/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})*({pd*pb}x^{pb-1}))/({-pc*pa})}\\$',if(mod(pb,2)=0,infinity,-infinity)]\n\n\n]\n", "name": "power_log", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
"}, "nb": {"templateType": "anything", "definition": "nl[1]", "name": "nb", "group": "Ungrouped variables", "description": ""}, "pL": {"templateType": "anything", "definition": "shuffle(1..12)[0..5]", "name": "pL", "group": "Ungrouped variables", "description": ""}, "na": {"templateType": "anything", "definition": "nl[0]", "name": "na", "group": "Ungrouped variables", "description": ""}, "exp_log": {"templateType": "anything", "definition": "[\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pd}e^({pa}x+{nb})+{nc})/({pc}ln({pb}x+{nb}))}\\$','\\$\\\\simplify{({pd}*{pa}*e^({pa}x+{nb}))/({pc*pb}/({pb}x+{nb}))=({pd}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))/{pc*pb}}\\$',infinity], \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pd}e^({pa}x+{nb})+{nc})/({pc}ln({pb}x+{nb}))}\\$','\\$\\\\simplify{({-pd}*{pa}*e^({pa}x+{nb}))/({pc*pb}/({pb}x+{nb}))=({-pd}*{pa}*e^({pa}x+{nb})*({pb}x+{nb}))/{pc*pb}}\\$',-infinity], \n\n\n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({pd}e^({-pa}x+{nb})+{nc})/({pc}ln({-pb}x+{nb}))}\\$','\\$\\\\simplify{({pd}*{-pa}*e^({-pa}x+{nb}))/({pc*-pb}/({-pb}x+{nb}))=({pd}*{-pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))/{pc*-pb}}\\$',infinity],\n ['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$', '\\$\\\\simplify{({-pd}e^({-pa}x+{nb})+{nc})/({pc}ln({-pb}x+{nb}))}\\$','\\$\\\\simplify{({-pd}*{-pa}*e^({-pa}x+{nb}))/({pc*-pb}/({-pb}x+{nb}))=({pd}*{pa}*e^({-pa}x+{nb})*({-pb}x+{nb}))/{pc*-pb}}\\$',-infinity]\n]", "name": "exp_log", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
"}, "pe": {"templateType": "anything", "definition": "pl[4]", "name": "pe", "group": "Ungrouped variables", "description": ""}, "nL": {"templateType": "anything", "definition": "shuffle(-12..12 except 0)[0..5]", "name": "nL", "group": "Ungrouped variables", "description": ""}, "choice1": {"templateType": "anything", "definition": "random(0..7)", "name": "choice1", "group": "Ungrouped variables", "description": ""}, "root_log": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({pa}x+{nb},{pb}))/({pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))/({pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})^(1/{pb}))/({pc*pb})}\\$',infinity], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({-pa}x+{nb},{pb}))/({pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))/({-pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})^(1/{pb}))/({pc*pb})}\\$',infinity], \n \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({pa}x+{nb},{pb}))/({-pc}ln({pa}x+{nb}))}\\$','\\$\\\\simplify{({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))/({-pc*pa}/({pa}x+{nb}))=(({pa}x+{nb})^(1/{pb}))/({-pc*pb})}\\$',-infinity], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{(root({-pa}x+{nb},{pb}))/({-pc}ln({-pa}x+{nb}))}\\$','\\$\\\\simplify{({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))/({pc*pa}/({-pa}x+{nb}))=(({-pa}x+{nb})^(1/{pb}))/({-pc*pb})}\\$',-infinity]\n\n]", "name": "root_log", "group": "Ungrouped variables", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
"}, "nd": {"templateType": "anything", "definition": "nl[3]", "name": "nd", "group": "Ungrouped variables", "description": ""}, "log_root": {"templateType": "anything", "definition": "[\n\n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pc}ln({pa}x+{nb}))/(root({pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({pc*pa}/({pa}x+{nb}))/({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))=({pc*pb})/(({pa}x+{nb})^(1/{pb}))}\\$',0], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({pc}ln({-pa}x+{nb}))/(root({-pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({-pc*pa}/({-pa}x+{nb}))/({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))=({pc*pb})/(({-pa}x+{nb})^(1/{pb}))}\\$',0], \n \n['\\$x\\\\rightarrow\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pc}ln({pa}x+{nb}))/(root({pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({-pc*pa}/({pa}x+{nb}))/({pa}/{pb}*(({pa}x+{nb})^({1-pb}/{pb})))=({-pc*pb})/(({pa}x+{nb})^(1/{pb}))}\\$',0], \n['\\$x\\\\rightarrow-\\\\infty\\$','\\$\\\\frac{-\\\\infty}{\\\\infty}\\$','\\$\\\\simplify{({-pc}ln({-pa}x+{nb}))/(root({-pa}x+{nb},{pb}))}\\$','\\$\\\\simplify{({pc*pa}/({-pa}x+{nb}))/({-pa}/{pb}*(({-pa}x+{nb})^({1-pb}/{pb})))=({-pc*pb})/(({-pa}x+{nb})^(1/{pb}))}\\$',0]\n\n]", "name": "log_root", "group": "Ungrouped variables", "description": "using a list to keep track of important things
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"}}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "Just what the title says, I guess.
"}, "functions": {}, "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "statement": "This question is about limits of indeterminate forms.
", "type": "question"}, {"name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form 0/0", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "functions": {}, "parts": [{"scripts": {}, "type": "gapfill", "gaps": [{"checkingtype": "absdiff", "type": "jme", "showCorrectAnswer": false, "vsetrangepoints": 5, "showFeedbackIcon": true, "expectedvariablenames": [], "variableReplacements": [], "answer": "{{choice2}[-2]}", "variableReplacementStrategy": "originalfirst", "scripts": {}, "marks": 1, "showpreview": true, "checkvariablenames": false, "checkingaccuracy": "0.0000000001", "vsetrange": [0, 1]}], "showCorrectAnswer": true, "prompt": "As {choice2[0]}, {choice2[2]} approaches [[0]]
\nNote: infinity is simply entered by typing infinity
", "showFeedbackIcon": true, "marks": 0, "variableReplacements": [], "variableReplacementStrategy": "originalfirst"}], "tags": [], "advice": "As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)=0$, $\\lim_{x\\rightarrow a}g(x)=0$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:
\n\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]
\n(If this is still an indeterminate form we can (of course) repeat the procedure)
\n\nSo by applying L'Hospital's rule we now need to determine what {choice2[3]} approaches as {choice2[0]}. But this is again an indeterminate form so we repeatedly apply L'Hospital's rule until it is not an indeterminate form and we arrive at needing to determine what {choice2[4]} approaches as {choice2[0]}. Now that this is no longer an indeterminate form, it should be clear that the limit is {choice2[-1]}.
\n", "statement": "This question is about limits of indeterminate forms.
", "variable_groups": [], "preamble": {"css": "", "js": ""}, "rulesets": {}, "metadata": {"licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International", "description": "Just what the title says, I guess.
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", "group": "Ungrouped variables"}, "choice1": {"name": "choice1", "definition": "random(sin_poly,cos_poly,poly_sin,poly_cos,trig_trig,log_sin,log_cos,log_tan,sin_log,cos_log,tan_log)", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "pL": {"name": "pL", "definition": "shuffle(2..12)[0..5]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "ne": {"name": "ne", "definition": "nl[4]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "sin_poly": {"name": "sin_poly", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/((x-{na})*(x-{nb}))}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/(2x-{na+nb})}\\$',nc/(na-nb),'\\$\\\\simplify[fractionNumbers]{{nc/(na-nb)}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x^2-{na+nb}x+{na*nb})}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/(2x-{na+nb})}\\$',nc/(na-nb),'\\$\\\\simplify[fractionNumbers]{{nc/(na-nb)}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x-{na})^2}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/(2(x-{na}))}\\$',0,'\\$0\\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x))/(x^2-{nb}x)}\\$','\\$\\\\simplify{({nc}cos(x))/(2x-{nb})}\\$',-nc/nb,'\\$\\\\simplify[fractionNumbers]{{-nc/nb}}\\$'], \n['\\$x\\\\rightarrow 0^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x))/x^2}\\$','\\$\\\\simplify{({nc}cos(x))/(2x)}\\$',nc*infinity,'\\$\\\\simplify{{nc*infinity}}\\$'],\n['\\$x\\\\rightarrow 0^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x))/x^2}\\$','\\$\\\\simplify{({nc}cos(x))/(2x)}\\$',-nc*infinity,'\\$\\\\simplify{-{nc*infinity}}\\$'],\n \n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x-{na})^{pa}}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/({pa}(x-{na})^{pa-1})}\\$',nc*infinity,'\\$\\\\simplify{{nc*infinity}}\\$'],\n['\\$x\\\\rightarrow\\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}sin(x-{na}))/(x-{na})^{pa}}\\$','\\$\\\\simplify{({nc}cos(x-{na}))/({pa}(x-{na})^{pa-1})}\\$',if(mod(pa,2)=0,-nc*infinity,nc*infinity),if(mod(pa,2)=0,'\\$\\\\simplify{{-nc*infinity}}\\$','\\$\\\\simplify{{nc*infinity}}\\$')] \n\n]", "templateType": "anything", "description": "using a list to keep track of important things
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", "group": "Ungrouped variables"}, "choice2": {"name": "choice2", "definition": "random(choice1)", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "cos_log": {"name": "cos_log", "definition": "[\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}cos(x-{na})-{pa})/({nb}ln(x-{na-1}))}\\$','\\$\\\\simplify{((x-{na-1})*({-pa}sin(x-{na})))/({nb})}\\$',0,'\\$ 0\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}cos(x)-{pa})/({nb}ln(x+1))}\\$','\\$\\\\simplify{((x+1)*({-pa}sin(x)))/({nb})}\\$',0,'\\$ 0 \\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}cos(x-{na})-{pa})/({nb}ln(x/{na}))}\\$','\\$\\\\simplify{(x*({-pa}sin(x-{na})))/({nb})}\\$',0,'\\$ 0 \\$'], \n['\\$x\\\\rightarrow 1\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}cos(pi/2*x))/({nb}ln(x))}\\$','\\$\\\\simplify{((x)*({-pa}*pi*sin(pi/2*x)))/({2*nb})}\\$',(-pi*pa)/(2*nb),'\\$\\\\simplify[fractionNumbers,simplifyFractions,unitFactor,unitDenominator]{({-pa}*{pi})/{2*nb}}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}cos((x-{na})^2)-{pa})/({nb}ln(x-{na-1}))}\\$','\\$\\\\simplify{((x-{na-1})*{-pa}*2(x-{na})*sin((x-{na})^2))/({nb})}\\$',0,'\\$ 0\\$']\n]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "cos_poly": {"name": "cos_poly", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x-{na})-{nc})/((x-{na})*(x-{nb}))}\\$','\\$\\\\simplify{({-nc}sin(x-{na}))/(2x-{na+nb})}\\$',0,'\\$ 0\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x-{na})-{nc})/(x^2-{na+nb}x+{na*nb})}\\$','\\$\\\\simplify{({-nc}sin(x-{na}))/(2x-{na+nb})}\\$',0,'\\$ 0\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x-{na})-{nc})/(x-{na})^2}\\$','\\$\\\\simplify{({-nc}sin(x-{na}))/(2(x-{na}))}\\$','\\$\\\\simplify{({-nc}cos(x-{na}))/(2)}\\$',-nc/2,'\\$\\\\simplify[fractionNumbers]{{-nc/2}}\\$'],\n\n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x)-{nc})/(x^2-{nb}x)}\\$','\\$\\\\simplify{({-nc}sin(x))/(2x-{nb})}\\$',0,'\\$0\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x)-{nc})/x^2}\\$','\\$\\\\simplify{({-nc}sin(x))/(2x)}\\$','\\$\\\\simplify{({-nc}cos(x))/(2)}\\$',-nc/2,'\\$\\\\simplify[fractionNumbers]{{-nc/2}}\\$'],\n \n['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x-{na})-{nc})/(x-{na})^{pa}}\\$','\\$\\\\simplify{({-nc}sin(x-{na}))/({pa}(x-{na})^{pa-1})}\\$','\\$\\\\simplify[unitPower,zeroPower]{({-nc}cos(x-{na}))/({pa*(pa-1)}(x-{na})^{pa-2})}\\$',switch(pa=2,-nc/(pa*(pa-1)),-nc*infinity),switch(pa=2,'\\$\\\\var[fractionNumbers]{-nc/(pa*(pa-1))}\\$','\\$\\\\var{-nc*infinity}\\$')],\n['\\$x\\\\rightarrow\\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x-{na})-{nc})/(x-{na})^{pa}}\\$','\\$\\\\simplify{({-nc}sin(x-{na}))/({pa}(x-{na})^{pa-1})}\\$','\\$\\\\simplify[unitPower,zeroPower]{({-nc}cos(x-{na}))/({pa*(pa-1)}(x-{na})^{pa-2})}\\$',switch(pa=2,-nc/(pa*(pa-1)),mod(pa,2)=0,-nc*infinity,nc*infinity),switch(pa=2,'\\$\\\\var[fractionNumbers]{-nc/(pa*(pa-1))}\\$',mod(pa,2)=0,'\\$\\\\var{-nc*infinity}\\$','\\$\\\\var{nc*infinity}\\$')]\n]", "templateType": "anything", "description": "using a list to keep track of important things
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", "group": "Ungrouped variables"}, "log_sin": {"name": "log_sin", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}sin(x-{na}))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*({pa}cos(x-{na})))}\\$',nb/pa,'\\$\\\\simplify[fractionNumbers]{{nb/pa}}\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x+1))/({pa}sin(x))}\\$','\\$\\\\simplify{({nb})/((x+1)*({pa}cos(x)))}\\$',nb/pa,'\\$\\\\simplify[fractionNumbers]{{nb/pa}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x/{na}))/({pa}sin(x-{na}))}\\$','\\$\\\\simplify{({nb})/(x*({pa}cos(x-{na})))}\\$',nb/(pa*na),'\\$\\\\simplify[fractionNumbers,simplifyFractions]{{nb}/{pa*na}}\\$'], \n['\\$x\\\\rightarrow 1\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x))/({pa}sin(pi*x))}\\$','\\$\\\\simplify{({nb})/((x)*({pa}*pi*cos(pi*x)))}\\$',-nb/(pi*pa),'\\$\\\\simplify[fractionNumbers,simplifyFractions,unitFactor]{{-nb}/({pa}*{pi})}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}sin((x-{na})^2))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*{pa}*2(x-{na})*cos((x-{na})^2))}\\$',nb*infinity,'\\$\\\\simplify{{nb*infinity}}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}sin((x-{na})^2))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*{pa}*2(x-{na})*cos((x-{na})^2))}\\$',-nb*infinity,'\\$\\\\simplify{{-nb*infinity}}\\$']\n]", "templateType": "anything", "description": "using a list to keep track of important things
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", "group": "Ungrouped variables"}, "sin_log": {"name": "sin_log", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x-{na}))/({nb}ln(x-{na-1}))}\\$','\\$\\\\simplify{((x-{na-1})*({pa}cos(x-{na})))/({nb})}\\$',pa/nb,'\\$\\\\simplify[fractionNumbers]{{pa/nb}}\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x))/({nb}ln(x+1))}\\$','\\$\\\\simplify{((x+1)*({pa}cos(x)))/({nb})}\\$',pa/nb,'\\$\\\\simplify[fractionNumbers]{{pa/nb}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x-{na}))/({nb}ln(x/{na}))}\\$','\\$\\\\simplify{(x*({pa}cos(x-{na})))/({nb})}\\$',(pa*na)/nb,'\\$\\\\simplify[fractionNumbers,simplifyFractions,unitDenominator]{{pa*na}/{nb}}\\$'], \n['\\$x\\\\rightarrow 1\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(pi*x))/({nb}ln(x))}\\$','\\$\\\\simplify{((x)*({pa}*pi*cos(pi*x)))/({nb})}\\$',(-pi*pa)/nb,'\\$\\\\simplify[fractionNumbers,simplifyFractions,unitFactor,unitDenominator]{({-pa}*{pi})/{nb}}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin((x-{na})^2))/({nb}ln(x-{na-1}))}\\$','\\$\\\\simplify{((x-{na-1})*{pa}*2(x-{na})*cos((x-{na})^2))/({nb})}\\$',0,'\\$ 0\\$']\n]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "pb": {"name": "pb", "definition": "pl[1]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "nd": {"name": "nd", "definition": "nl[3]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "tan_log": {"name": "tan_log", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}tan(x-{na}))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*({pa}sec(x-{na})^2))=({nb}cos(x-{na})^2)/((x-{na-1})*{pa})}\\$',nb/pa,'\\$\\\\simplify[fractionNumbers]{{nb/pa}}\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x+1))/({pa}tan(x))}\\$','\\$\\\\simplify{({nb})/((x+1)*{pa}*sec(x)^2)=({nb}cos(x)^2)/((x+1)*{pa})}\\$',nb/pa,'\\$\\\\simplify[fractionNumbers]{{nb/pa}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x/{na}))/({pa}tan(x-{na}))}\\$','\\$\\\\simplify{({nb})/(x*{pa}*sec(x-{na})^2)=({nb}*cos(x-{na})^2)/(x*{pa})}\\$',nb/(pa*na),'\\$\\\\simplify[fractionNumbers,simplifyFractions]{{nb}/{pa*na}}\\$'], \n['\\$x\\\\rightarrow 1\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x))/({pa}tan(pi*x))}\\$','\\$\\\\simplify{({nb})/((x)*{pa}*pi*sec(pi*x)^2)=({nb}*cos(pi*x)^2)/((x)*{pa}*pi)}\\$',nb/(pi*pa),'\\$\\\\simplify[fractionNumbers,simplifyFractions,unitFactor]{{nb}/({pa}*{pi})}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}tan((x-{na})^2))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*{pa}*2(x-{na})*sec((x-{na})^2)^2)=({nb}*cos((x-{na})^2)^2)/((x-{na-1})*{pa}*2(x-{na}))}\\$',nb*infinity,'\\$\\\\simplify{{nb*infinity}}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}tan((x-{na})^2))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*{pa}*2(x-{na})*sec((x-{na})^2)^2)=({nb}*cos((x-{na})^2)^2)/((x-{na-1})*{pa}*2(x-{na}))}\\$',-nb*infinity,'\\$\\\\simplify{{-nb*infinity}}\\$']\n]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "trig_trig": {"name": "trig_trig", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x-{na})-{nc})/({pa}sin(x-{na}))}\\$','\\$\\\\simplify{({-nc}sin(x-{na}))/({pa}cos(x-{na}))}\\$',0,'\\$ 0\\$'], \n\n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nc}cos(x)-{nc})/({pa}sin(x))}\\$','\\$\\\\simplify{({-nc}sin(x))/({pa}cos(x))}\\$',0,'\\$0\\$'], \n\n ['\\$x\\\\rightarrow\\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x-{na}))/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{({pa}cos(x-{na}))/({-nc}sin(x-{na}))}\\$',(-pa/nc)*infinity,'\\$ \\\\simplify{{(-pa/nc)*infinity}}\\$'], \n ['\\$x\\\\rightarrow\\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x-{na}))/({nc}cos(x-{na})-{nc})}\\$','\\$\\\\simplify{({pa}cos(x-{na}))/({-nc}sin(x-{na}))}\\$',(pa/nc)*infinity,'\\$ \\\\simplify{{(pa/nc)*infinity}}\\$'],\n \n['\\$x\\\\rightarrow 0^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x))/({nc}cos(x)-{nc})}\\$','\\$\\\\simplify{({pa}cos(x))/({-nc}sin(x))}\\$',(-pa/nc)*infinity,'\\$ \\\\simplify{{(-pa/nc)*infinity}}\\$'],\n['\\$x\\\\rightarrow 0^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({pa}sin(x))/({nc}cos(x)-{nc})}\\$','\\$\\\\simplify{({pa}cos(x))/({-nc}sin(x))}\\$',(pa/nc)*infinity,'\\$ \\\\simplify{{(pa/nc)*infinity}}\\$']\n]", "templateType": "anything", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
", "group": "Ungrouped variables"}, "log_tan": {"name": "log_tan", "definition": "[\n\n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}tan(x-{na}))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*({pa}sec(x-{na})^2))=({nb}cos(x-{na})^2)/((x-{na-1})*{pa})}\\$',nb/pa,'\\$\\\\simplify[fractionNumbers]{{nb/pa}}\\$'], \n['\\$x\\\\rightarrow 0\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x+1))/({pa}tan(x))}\\$','\\$\\\\simplify{({nb})/((x+1)*{pa}*sec(x)^2)=({nb}cos(x)^2)/((x+1)*{pa})}\\$',nb/pa,'\\$\\\\simplify[fractionNumbers]{{nb/pa}}\\$'], \n['\\$x\\\\rightarrow\\\\var{na}\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x/{na}))/({pa}tan(x-{na}))}\\$','\\$\\\\simplify{({nb})/(x*{pa}*sec(x-{na})^2)=({nb}*cos(x-{na})^2)/(x*{pa})}\\$',nb/(pa*na),'\\$\\\\simplify[fractionNumbers,simplifyFractions]{{nb}/{pa*na}}\\$'], \n['\\$x\\\\rightarrow 1\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x))/({pa}tan(pi*x))}\\$','\\$\\\\simplify{({nb})/((x)*{pa}*pi*sec(pi*x)^2)=({nb}*cos(pi*x)^2)/((x)*{pa}*pi)}\\$',nb/(pi*pa),'\\$\\\\simplify[fractionNumbers,simplifyFractions,unitFactor]{{nb}/({pa}*{pi})}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}^+\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}tan((x-{na})^2))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*{pa}*2(x-{na})*sec((x-{na})^2)^2)=({nb}*cos((x-{na})^2)^2)/((x-{na-1})*{pa}*2(x-{na}))}\\$',nb*infinity,'\\$\\\\simplify{{nb*infinity}}\\$'],\n['\\$x\\\\rightarrow \\\\var{na}^-\\$','\\$\\\\frac{0}{0}\\$','\\$\\\\simplify{({nb}ln(x-{na-1}))/({pa}tan((x-{na})^2))}\\$','\\$\\\\simplify{({nb})/((x-{na-1})*{pa}*2(x-{na})*sec((x-{na})^2)^2)=({nb}*cos((x-{na})^2)^2)/((x-{na-1})*{pa}*2(x-{na}))}\\$',-nb*infinity,'\\$\\\\simplify{{-nb*infinity}}\\$']\n]", "templateType": "anything", "description": "using a list to keep track of important things
\nwhat x approaches, indeterminate form type, what y is, the first application of L'Hospital's, (if needed) the final application of L'Hopital's, what y approaches.
", "group": "Ungrouped variables"}, "nc": {"name": "nc", "definition": "nl[2]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}, "nL": {"name": "nL", "definition": "shuffle(-12..12 except 0)[0..5]", "templateType": "anything", "description": "", "group": "Ungrouped variables"}}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "alt1", "sin_poly", "cos_poly", "poly_sin", "poly_cos", "trig_trig", "log_sin", "log_cos", "log_tan", "sin_log", "cos_log", "tan_log", "choice1", "choice2"], "variablesTest": {"maxRuns": 100, "condition": ""}, "type": "question"}, {"name": "Maria's copy of Limits: L'Hospital's rule: Indeterminate form 0*infinity", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}], "advice": "As {choice2[0]}, it appears {choice2[2]} approaches {choice2[1]}, but it isn't clear what this means. This is known as an indeterminate form of type {choice2[1]}. We rewrite this product as a quotient {choice2[4]} so that it is an indeterminate form of type {choice2[3]}. L'Hospital's rule says that if we have such an indeterminate form we can differentiate the numerator and denominator separately and take the limit of the quotient. That is, if $f$ and $g$ are differentiable on an open interval containing $a$, $g'(x)\\ne 0$ on that interval (except possibly at $x=a$), $\\lim_{x\\rightarrow a}f(x)$ and $\\lim_{x\\rightarrow a}g(x)$ both equal $\\infty$ or both equal $0$ and $\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}$ exists (or equals $\\pm\\infty)$ then:
\n\\[\\lim_{x\\rightarrow a} \\frac{f(x)}{g(x)}=\\lim_{x\\rightarrow a}\\frac{f'(x)}{g'(x)}.\\]
\n(If this is still an indeterminate form we can (of course) repeat the procedure)
\n\nSo by applying L'Hospital's rule we now need to determine what {choice2[5]} approaches as {choice2[0]}. But this is again an indeterminate form so we repeatedly apply L'Hospital's rule until it is not an indeterminate form and we arrive at needing to determine what {choice2[6]} approaches as {choice2[0]}. Now that this is no longer an indeterminate form, it should be clear that the limit is {choice2[-1]}.
\n", "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["pL", "pa", "pb", "pc", "pd", "pe", "nL", "na", "nb", "nc", "nd", "ne", "alt1", "power_exp", "log_power", "exp_power", "log_sin", "log_tan", "tan_linear", "choice1", "choice2"], "parts": [{"type": "gapfill", "showCorrectAnswer": true, "variableReplacements": [], "gaps": [{"expectedvariablenames": [], "checkingtype": "absdiff", "vsetrangepoints": 5, "vsetrange": [0, 1], "scripts": {}, "answer": "{{choice2}[-2]}", "variableReplacementStrategy": "originalfirst", "showpreview": true, "type": "jme", "showCorrectAnswer": false, "checkingaccuracy": "0.0000000001", "showFeedbackIcon": true, "checkvariablenames": false, "variableReplacements": [], "marks": 1}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "marks": 0, "prompt": "As {choice2[0]}, {choice2[2]} approaches [[0]]
\nNote: infinity is simply entered by typing infinity
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\n\\(x_1=\\) [[0]]
\n\\(x_2=\\) [[1]]
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\n\\(x_4=\\) [[3]]
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\n\\(f(x)=\\var{a}x^3-\\var{b}x^2+\\var{c}x+\\var{d}\\)
\nGive your answers correct to three decimal places.
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